Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $t = \dfrac{z^2 - 12z + 27}{2z - 18} \times \dfrac{z + 8}{6z - 18} $
Solution: First factor the quadratic. $t = \dfrac{(z - 3)(z - 9)}{2z - 18} \times \dfrac{z + 8}{6z - 18} $ Then factor out any other terms. $t = \dfrac{(z - 3)(z - 9)}{2(z - 9)} \times \dfrac{z + 8}{6(z - 3)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ (z - 3)(z - 9) \times (z + 8) } { 2(z - 9) \times 6(z - 3) } $ $t = \dfrac{ (z - 3)(z - 9)(z + 8)}{ 12(z - 9)(z - 3)} $ Notice that $(z - 9)$ and $(z - 3)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ \cancel{(z - 3)}(z - 9)(z + 8)}{ 12(z - 9)\cancel{(z - 3)}} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $t = \dfrac{ \cancel{(z - 3)}\cancel{(z - 9)}(z + 8)}{ 12\cancel{(z - 9)}\cancel{(z - 3)}} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $t = \dfrac{z + 8}{12} ; \space z \neq 3 ; \space z \neq 9 $